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Two positive charges of magnitude $q$ are placed at the ends of a side $1$ of a square of side $2a$. Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge $Q$, moves from the middle of side $1$ to the centre of square, its kinetic energy at the centre of square is
$\frac{1}{{4\pi \,\,{\varepsilon_{0}}}}\,\frac{{2qQ}}{a}\,\left( {1\,\, - \,\frac{1}{{\sqrt 5 }}} \right)$
$zero$
$\frac{1}{{4\pi \,\,{\varepsilon_{0}}}}\,\frac{{2qQ}}{a}\,\,\left( {1\,\, + \,\,\frac{1}{{\sqrt 5 }}} \right)$
$\frac{1}{{4\pi \,\,{ \varepsilon_{0}}}}\,\frac{{2qQ}}{a}\,\left( {1\,\, - \,\frac{2}{{\sqrt 5 }}} \right)$
Solution
Initial potential of the charge,
$V_{A}=\frac{2 k q}{a}-\frac{2 k q}{a \sqrt{5}}$
$\Rightarrow V_{A}=\frac{1}{4 \pi E} \frac{2 q}{a}\left(1-\frac{1}{\sqrt{5}}\right)$
(Here potential due to each $q=\frac{k q}{a}$ and potential due to each $-q=\frac{-k q}{a \sqrt{5}}$)
Final potential of the charge
$V_{B}=0$
( $\because$ Point $B$ is equidistant from all the four charges)
$\therefore$ Using work energy theorem,
$\left(W_{A B}\right)_{\text {electric }}=Q\left(V_{A}-V_{B}\right)$
$=\frac{2 q Q}{4 \pi E_{0} a}\left[1-\frac{1}{\sqrt{5}}\right]$
$=\left(\frac{1}{4 \pi \varepsilon_{0}}\right) \frac{2 Q q}{a}\left[1-\frac{1}{\sqrt{5}}\right]$
